<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\pa}{\partial}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\begin{document}
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\begin{center}
\textbf{Vector Fields}

\textit{\textbf{Conservative Fields}}
\end{center}

\textbf{Question}

\begin{description}
\item{(a)}
Show that the gradient of a function which is expressed in polar
coordinates in the plane is
$$\nabla \phi (t, \theta) = \frac{\partial \phi}{\partial r}
\un{\hat{r}} + \frac{1}{r} \frac{\partial \phi}{\partial \theta}
\un{\hat{\theta}}.$$

\item{(b)}
Use the result from part (a) to show that a necessary condition for
the vector field
$$\un{F}(r, \theta) = F_r(r, \theta) \un{\hat{r}} + F_{\theta} (r,
\theta) \un{\hat{\theta}}$$
(expressed in polar coordinates) to be conservative is that
$$\frac{\partial F_r}{\partial \theta} - r \frac{\partial
F_{\theta}}{\partial r} = F_{\theta}.$$
\end{description}


\textbf{Answer}

\begin{description}
\item{(a)}
As $x=r\cos\theta$ and $y=r\sin\theta$
\begin{eqnarray*}
\frac{\pa \phi}{\pa r} &amp; = &amp; \cos \theta \frac{\pa \phi}{\pa x} +
\sin\theta \frac{\pa \phi}{\pa y}\\
\frac{\pa \phi}{\pa \theta} &amp; = &amp; -r\sin\theta \frac{\pa \phi}{\pa x}
+ r \cos \theta \frac{\pa\phi}{\pa y}\\
\textrm{And} &amp; &amp; \\
\un{\hat{r}} &amp; = &amp; \frac{x\un{i} + y \un{j}}{r} = (\cos\theta )\un{i}
+ (\sin\theta)\un{j}\\
\un{\hat{\theta}} &amp; = &amp; \frac{-y\un{i} + x\un{j}}{r} =
-(\sin\theta)\un{i} + (\cos\theta)\un{j}
\end{eqnarray*}
This leads to the fact that
\begin{eqnarray*}
\frac{\pa\phi}{\pa r} \un{\hat{r}} + \frac{1}{r} \frac{\pa\phi}{\pa
\theta} \un{\hat{\theta}} &amp; = &amp;
\left ( \cos^2\theta \frac{\pa\phi}{\pa x} + \sin\theta\cos\theta
\frac{\pa\phi}{ \pa y} \right ) \un{i}\\
&amp; &amp; + \left (\cos\theta\sin\theta \frac{\pa\phi}{\pa x} + \sin^2\theta
\frac{\pa \phi}{\pa y} \right ) \un{j}\\
&amp; &amp; + \left ( \sin^2 \theta \frac{\pa \phi}{\pa x} - \sin\theta\cos\theta
\frac{\pa\phi }{\pa y} \right ) \un{i}\\
&amp; &amp; + \left ( -\cos\theta\sin\theta\frac{\pa\phi}{\pa x} +
\cos^2\theta \frac{\pa\phi}{\pa y} \right ) \un{j}\\
&amp; = &amp; \frac{\pa \phi}{\pa x} \un{i} + \frac{\pa \phi}{\pa y} \un{j} =
\nabla \phi.
\end{eqnarray*}  

\item{(b)}
If $\un{F}(r, \theta) = F_r(r, \theta) \un{\hat{r}} + F_{\theta} (r,
\theta) \un{\hat{\theta}}$ is conservative, then $\un{F} = \nabla
\phi$ for a scalar field $\phi(r, \theta)$. Using part (a)
$$\frac{\pa\phi}{\pa r}=F_r, \ \ \ \frac{1}{r}\frac{\pa\phi}{\pa
\theta}=F_{\theta}.$$
In order to have mixed second partial derivatives of $\phi$ that are
equal, it is necessary to have
$$\frac{\pa F_r}{\pa \theta} = \frac{\pa}{\pa r}(rF_{\theta}) =
F_{\theta} + r\frac{\pa F_{\theta}}{\pa r}$$
$$\textrm{i.e. }\frac{\partial F_r}{\partial \theta} - r \frac{\partial
F_{\theta}}{\partial r} = F_{\theta}.$$
\end{description}

\end{document}




















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